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(B) Let $T$ be the tension in the rope and $N$ be the normal force between the man and the platform. The man and the platform move together with an up

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(B) Let $T$ be the tension in the rope and $N$ be the normal force between the man and the platform. The man and the platform move together with an upward acceleration $a = 2 \ m/s^2$.For the man:$T + N - m_m g = m_m a$$T + N - 60(10) = 60(2)$$T + N = 600 + 120 = 720 \quad ...(1)$For the platform:There are two segments of the rope pulling the platform upward (one from the fixed pulley and one from the movable pulley system). The total upward force on the platform is $2T$. $2T - N - m_p g = m_p a$$2T - N - 40(10) = 40(2)$$2T - N = 400 + 80 = 480 \quad ...(2)$Adding equations $(1)$ and $(2)$:$(T + N) + (2T - N) = 720 + 480$$3T = 1200$$T = 400 \ N$Thus,the man must apply a force of $400 \ N$ on the rope.

$A$ man of mass $60 \ kg$ is standing on a platform of mass $40 \ kg$ as shown in the figure. What force should the man apply on the rope so that he accelerates upward with the platform with an acceleration of $2 \ m/s^2$ (in $N$)? (Take $g = 10 \ m/s^2$)

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